Go To Your Study Dashboard

Get Answers to all your Questions

header-bg

For each t $\epsilon$ R, let [t] be the greatest integer less than or equal to t. Then

$\lim_{x\rightarrow 0}x\left ( \left [1/x\right ]\left +[2/x \right ]\left+ .....+[15/x \right ] \right )$

Option 1)
does not exist (in R).

Option 2)
is equal to 0.

Option 3)
is equal to 15

Option 4)
is equal to 120

Answers (2)

best_answer

As we learnt in

The SANDWICH THEOREM -

$If \:\:f(x)\leq g(x)\leq h(x)\:\:for\:every\:(x)\:in\:the\:deleted\:neighbourhood\:of\:(a).$

$and\:\:\:\lim_{x\rightarrow a}\:\:f(x)=\lim_{x\rightarrow a}\:\:h(x)=l$

$Then\;\:\lim_{x\rightarrow a}\:g(x)\:\:is\:also\:equal\:to\:l$

- wherein

$Where\:\:\:x\epsilon \:(a-\delta ,\:a+\delta )\:\:and\:\:\delta\: \:is\:very\:small.$

$x-1 < \left [ x \right ]\leq x$

$1/x-1 < \left [ 1/x \right ]\leq 1/x$

$2/x-1 < \left [ 2/x \right ]\leq 2/x$

.

.

.

.

$15/x-1 < \left [ 15/x \right ]\leq 15/x$

ADD ALL ;

$120/x- 15< f(n)\leq 120/x$

$120-15x/x< f(n)\leq 120/x$

$\lim_{x\rightarrow 0^{+}} (\frac{120-15x}{x})< \lim_{x\rightarrow 0^{+}}f(x)\leq \lim_{x\rightarrow 0^{+} } 120/x$

$\lim_{x\rightarrow 0^{+}} 120 -15x< \lim_{x\rightarrow 0^{+}}f(x)\leq 120$

Thus $\lim_{x\rightarrow 0}f(x)=120$

Option 1)

does not exist (in R).

Option 2)

is equal to 0.

Option 3)

is equal to 15

Option 4)

is equal to 120

Posted by

Himanshu

View full answer

JEE Main high-scoring chapters and topics

Study 40% syllabus and score up to 100% marks in JEE

Similar Questions

Trending Articles/News

anam-khan

JEE Main 2025: BTech in AI gains popularity; IIT, NIT cut-offs for artificial intelligence

anam-khan

JEE Main 2025 Live: NTA JEE session 1 begins tomorrow; admit card link, exam pattern, passing marks

anam-khan

JEE Main 2025 exam centre changed at UAE's Sharjah due to “unavoidable circumstances”

Latest Question